Question:

Step 1: Find intersection with X-axis The X-axis is defined by $y=0$. Substitute $y=0$ into the curve equation $x+y=xy$: $$x + 0 = x(0)$$ $$x = 0$$ Thus, the curve intersects the X-axis at the point $(0,0)$. Step 2: Find $\frac{dy}{dx}$ using implicit differentiation The given equation is $x+y=xy$. Differentiate both sides with respect to $x$: $$\frac{d}{dx}(x+y) = \frac{d}{dx}(xy)$$ $$1 + \frac{dy}{dx} = y \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(y)$$ $$1 + \frac{dy}{dx} = y(1) + x\frac{dy}{dx}$$ $$1 + \frac{dy}{dx} = y + x\frac{dy}{dx}$$ Rearrange to solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} - x\frac{dy}{dx} = y - 1$$ $$\frac{dy}{dx}(1 - x) = y - 1$$ $$\frac{dy}{dx} = \frac{y - 1}{1 - x}$$ Step 3: Find slope of tangent at $(0,0)$ Substitute the coordinates of the intersection point $(0,0)$ into the expression for $\frac{dy}{dx}$: $$\frac{dy}{dx}\Big|_{(0,0)} = \frac{0 - 1}{1 - 0} = \frac{-1}{1} = -1$$ The slope of the tangent at $(0,0)$ is $-1$. Step 4: Find slope of normal The slope of the normal ($m_N$) is the negative reciprocal of the slope of the tangent ($m_T$): $$m_N = -\frac{1}{m_T} = -\frac{1}{-1} = 1$$ The slope of the normal is $1$. Step 5: Write equation of normal Using the point-slope form of a line, $y - y_1 = m(x - x_1)$, with the point $(0,0)$ and slope $m=1$: $$y - 0 = 1(x - 0)$$ $$y = x$$ This equation can be rewritten as $x - y = 0$.